# Lifting back homomorphisms from localized modules

Refer to exercises 9, 10 of chapter 3 in Lang's algebra, page 167.

In particular, let $A$ be a commutative ring, $p$ a prime ideal and $M, N$ $A$-modules. Then $M_p, N_p$ are the localized $A_p$ modules. Let $f: M_p \rightarrow N_p$ be a homomorphism of $A_p$ modules. How can we lift this homomorphism to an $A$-homomorphism $M \rightarrow N$ of the initial modules?

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Dear Manos, if you refer to an exercise in Lang, you should copy it in your question and not force users to try to get hold of that book. – Georges Elencwajg Oct 31 '11 at 9:42

You are starting with a sequence of $A$-modules $$0 \to M' \stackrel f\to M \stackrel g\to M'' \to 0.$$ You want to show that if for each prime $\mathfrak{p}$ of $A$ the induced sequence of $A_\mathfrak{p}$-modules $$0 \to M'_\mathfrak{p} \stackrel{f_\mathfrak{p}}\longrightarrow M_\mathfrak{p} \stackrel{g_\mathfrak{p}}\longrightarrow M''_\mathfrak{p} \to 0$$ is exact, then the original sequence is exact. Here the map $f_\mathfrak{p}$, for example, just sends $x/s$ to $f(x)/s$. You don't need to perform any sort of "lifting" on $f_\mathfrak{p}$, because you already have $f$. One way of solving the problem is to find ways to apply the fact that if $N_\mathfrak{p} = 0$ for all primes $\mathfrak{p}$, then $N = 0$. [You could try this with $N = \operatorname{Ker} f$. What is the relationship between $\operatorname{Ker} f$ and $\operatorname{Ker} f_\mathfrak{p}$?]

I should mention that if $M'$ is finitely presented then there is a relationship between $A$-linear $M' \to M$ and $S^{-1}A$-linear $S^{-1}M' \to S^{-1}M$. Proposition 2.10 of Eisenbud's book shows that in this situation we have a natural isomorphism $$\operatorname{Hom}_{S^{-1}A}(S^{-1}M', S^{-1}M) \approx S^{-1}\operatorname{Hom}_A(M', M).$$ So you might suggestively call Georges' example "the identity map divided by $2$".

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What if i start with a sequence $0 \rightarrow M_{p,1} \rightarrow M_{p,2} \rightarrow M_{p,3} \rightarrow 0$? Can i get a sequence $M_1 \rightarrow M_2 \rightarrow M_3$? Note that at this point i only want to get the sequence, i am not worrying about exactness. – Manos Oct 31 '11 at 3:00
@Manos I've no idea (though I think it would be best to just talk about one homomorphism $M'_\mathfrak{p} \to M_\mathfrak{p}$ then, since that doesn't seem any easier/harder). I'm most worried about the case when, say, $M$ doesn't inject into $M_\mathfrak{p}$. So I'd like to play around with examples like that and try to think about the problem geometrically. But your best bet is probably Georges :) – Dylan Moreland Oct 31 '11 at 3:31
Dear Manos, you have had two answers to your original question. If this does not satisfy you, I suggest you write another, precise question with quantificators in front of $p$. In particular the example in my post shows that the answer to your question above is "no" since if you cannot lift $0 \to M_p \to N_p \to 0$, you certainly cannot lift $0 \to M_p \to N_p \to 0_p \to 0$ – Georges Elencwajg Oct 31 '11 at 10:03
@Dylan: thanks Dylan, you revealed the point that was confusing me: i already have $f$, no need to do any lifting. – Manos Oct 31 '11 at 13:46
Since no one has mentioned it, perhaps it is worth noting that the functor $M\rightarrow M_p$ is the tensor product $\otimes_A A_p$. So we know certain things about it on more-general principles... – paul garrett Oct 31 '11 at 13:48

In general, given an $A_p$-linear morphism $f:M_p\to N_p$ it is impossible to find a morphism of $A$-modules $g: M\to N$ lifting $f$ in the sense that $g_p=f.$

For example, let $A,M$ and $N$ all be equal to $\mathbb Z$. Choose for $p$ the zero ideal $p=(0)$.
Then $A_p=M_p=N_p=\mathbb Q$ and you can choose for $f:\mathbb Q \to \mathbb Q$ the morphism $f(q)=\frac{q}{2}$.
The only $\mathbb Z$-linear maps $g:\mathbb Z \to \mathbb Z$ are of the form $g(n)=an$ for some $a\in \mathbb Z$ and clearly none of them lifts $f$.

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The issue is that the exercise 10b) says to prove that the sequence $0 \rightarrow M_1 \rightarrow M_2 \rightarrow M_3 \rightarrow 0$ is exact if and only if the sequence $0 \rightarrow M_{p,1} \rightarrow M_{p,2} \rightarrow M_{p,3} \rightarrow 0$ is exact. The forward direction is not a problem. The backward direction however requires some kind of lifting of the sequence to the initial modules. – Manos Oct 31 '11 at 0:27
@Manos: Take A=Z and p=0 as Georges suggests, then take M1 = 0, M2=Z/2Z, M3=0. Then the original sequence is not exact, but it becomes exact after localizing at p (it becomes 0→0→0→0→0). Presumably you need some extra hypotheses. – Jack Schmidt Oct 31 '11 at 1:23
@Manos You really want to require that the second sequence is exact for all primes $\mathfrak{p}$ of $A$. – Dylan Moreland Oct 31 '11 at 1:32
@Dylan: you are right, this is it. However, the question of how to lift back the sequence still remains. – Manos Oct 31 '11 at 1:59
@Manos Oh, do you want conditions under which a map $M_\mathfrak{p} \to N_\mathfrak{p}$ comes from an $M \to N$? That seems interesting. – Dylan Moreland Oct 31 '11 at 2:29